【数学A】例題2.2.14：ベイズの定理（One More）★★★★

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% 例題A2.2.14：ベイズの定理（One More）★★★★
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3つの箱A，B，Cがあり，箱Aには赤玉2個と白玉6個，箱Bには赤玉3個と白玉3個，箱Cには赤玉4個と白玉8個が入っている．3つの箱のうち1つを無作為に選び，その箱から1個の玉を取り出したところ赤玉であった．このとき，その赤玉が箱Aから取り出された玉である確率を求めよ．

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% 例題A2.2.14の解答（One More）★★★★
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箱Aを選ぶ事象を$A$，箱Bを選ぶ事象を$B$，箱Cを選ぶ事象を$C$，赤玉を取り出す事象を$R$とする．

箱A，箱B，箱Cを選ぶ確率$P(A),P(B),P(C)$は，すべて$\frac{1}{3}$

したがって，赤玉を取り出す確率は，

\begin{align*}
P(R)&=P(A\cap R)+P(B\cap R)+P(C\cap R)\\
&=P(A)P_A(R)+P(B)P_B(R)+P(C)P_C(R)\\
&=\frac{1}{3}\times\frac{2}{8}+\frac{1}{3}\times\frac{3}{6}+\frac{1}{3}\times\frac{4}{12}=\frac{13}{36}
\end{align*}

よって，求める確率は，

\[
P_R(A)=\frac{P(R\cap A)}{P(R)}=\frac{P(A\cap R)}{P(R)}=\frac{P(A)P_A(R)}{P(R)}=\frac{1}{12}\div\frac{13}{36}=\frac{3}{13}
\]

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% 問題A2.2.14：ベイズの定理（One More）★★★★
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3つの袋A，B，Cがあり，袋Aには赤いボール3個と青いボール5個，袋Bには赤いボール2個と青いボール6個，袋Cには赤いボール4個と青いボール4個が入っている．3つの袋のうち1つを無作為に選び，その袋から1個のボールを取り出したところ赤いボールであった．このとき，その赤いボールが袋Bから取り出されたものである確率を求めよ．

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% 問題A2.2.14の解答（One More）★★★★
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袋Aを選ぶ事象を$A$，袋Bを選ぶ事象を$B$，袋Cを選ぶ事象を$C$，赤いボールを取り出す事象を$R$とする．

袋A，袋B，袋Cを選ぶ確率$P(A),P(B),P(C)$は，すべて$\frac{1}{3}$

それぞれの袋から赤いボールを取り出す条件付き確率は，

\[
P_A(R)=\frac{3}{8},P_B(R)=\frac{2}{8},P_C(R)=\frac{4}{8}
\]

したがって，赤いボールを取り出す確率は，

\begin{align*}
P(R)&=P(A\cap R)+P(B\cap R)+P(C\cap R)\\
&=P(A)P_A(R)+P(B)P_B(R)+P(C)P_C(R)\\
&=\frac{1}{3}\times\frac{3}{8}+\frac{1}{3}\times\frac{2}{8}+\frac{1}{3}\times\frac{4}{8}=\frac{3}{8}
\end{align*}

よって，求める確率は，

\[
P_R(B)=\frac{P(B\cap R)}{P(R)}=\frac{P(B)P_B(R)}{P(R)}=\frac{1}{12}\div\frac{3}{8}=\frac{2}{9}
\]

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% 例題A2.2.14の側注：ベイズの定理（One More）★★★★
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% 側注1
乗法定理を用いる．

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% 問題A2.2.14の側注：ベイズの定理（One More）★★★★
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% 側注1
乗法定理を用いる．

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% 例題A2.2.14の考え方・補足など：ベイズの定理（One More）★★★★
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% 考え方
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下の余談におけるベイズの定理の，単純な場合である．箱Aを選ぶ事象を$A$，赤玉を取り出す事象を$R$とすると，求める確率は，条件付き確率$P_R(A)=\frac{P(R\cap A)}{P(R)}=\frac{P(A\cap R)}{P(R)}$である．そこで，$P(R),P(A\cap R)$を求めることを考える．

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% 余談
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上の例題から，$P_R(A)=\frac{P(A)P_A(R)}{P(A)P_A(R)+P(B)P_B(R)+P(C)P_C(R)}$が成り立つ．一般に，$A_1,A_2,\ldots,A_n$を互いに排反な事象とし，そのうちのいずれか1つが必ず起こるものとする．このとき，事象$B$が起こったという条件のもとで，事象$A_k$が起こる確率は，

\[
P_B(A_k)=\frac{P(A_k)P_{A_k}(B)}{P(A_1)P_{A_1}(B)+P(A_2)P_{A_2}(B)+\cdots+P(A_n)P_{A_n}(B)}(k=1,2,\ldots,n)
\]

である．これを，ベイズの定理という．